Probability it came from Bag A

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The discussion focuses on calculating probabilities related to selecting balls from two bags. Bag A contains 5 white and 2 green balls, while Bag B contains 3 white and 4 green balls. The probability of selecting a white ball is determined to be 11/21. For the conditional probability of a white ball coming from Bag A, the formula P(A|B) = P(A ∩ B) / P(B) is applied, confirming that this is indeed a conditional probability problem.

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Bag A contains 5 white balls and 2 green balls. Bag B contains 3 white balls and 4 green
balls. A fair die is rolled and if a 1 or a 2 comes up, a ball is randomly selected from Bag A;
however, if a 3, 4, 5 or 6 comes up, a ball is randomly selected from Bag B.
a) What is the probability of selecting a white ball?
b) If a white ball is selected, what is the probability that this ball came form Bag A?

Alright so a) I figured out. It's 11/21. I got that by multiplying the likelihood of first rolling a one or two (2/6) with the likelihood of getting a white ball in bag a (5/7) and then adding that number to this same process in bag b.

However for part b), for some reason I just can't wrap my head around it. I know that the probability of selecting a white ball is now 11/21, so is this an equation dealing with P(bag a ("given that" line)white ball probability)? Conditional probability? A little lost.
 
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Yes it is a conditional probability problem. You can use:

<br /> P(A|B) = \frac{P(A \cap B)}{P(B)}<br />

Where A and B are two events.
 
You found the chance of getting a white ball from bag a (Ill call it A), and you found the chance of getting a white ball in general (Ill call it B).
I would just divide A by B to get the probability that a white ball comes from bag A.
 

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